Wait—perhaps the problem allows degree ≤ 3. Many contest problems phrase cubic loosely. Given the values fit a quadratic, and no higher-degree terms are forced, the minimal-degree interpolating polynomial is quadratic. Since the problem asks to find $ p(0) $, and the unique cubic polynomial (in degree ≤ 3) satisfying the values must have $ a = 0 $, we proceed with $ p(x) = 2x^2 + x $, so $ p(0) = 0 $. However, to ensure degree 3, suppose we include a zero cubic term. Then $ p(x) = 0x^3 + 2x^2 + x + 0 $, and $ p(0) = 0 $. - Nelissen Grade advocaten